The magnitude of a velocity vector is called speed. suppose that a wind is blowing from the direction n45°w at a speed of 50 km/h.(this means that the direction from which the wind blows is 45° west of the northerly direction.) a pilot is steering a plane in the direction n60°e at an airspeed (speed in still air) of 200 km/h. the true course, or track, of the plane is the direction of the resultant of the velocity vectors of the plane and the wind. the ground speed of the plane is the magnitude of the resultant. find the true course and the ground speed of the plane.

give your answers correct to one decimal place.

true course in north of east
ground speed in km/h

To solve this problem, you must figure out (in vector form) both the wind vector and plane vector

w⃗ = wind vector

P⃗ = plane vector

 To get the  true course of the plane, you need to add the plane and wind vectors, the formula would be

w⃗ +P⃗ ,

 which will result to the ground speed.

ground speed=||w⃗ +P⃗ ||

Using the planar representation of your situation, this will help you understand the equation, use this to make the equation more understandable.

w⃗ =AB¯¯¯¯¯¯¯¯,   P⃗ =AC¯¯¯¯¯¯¯¯

the smaller circle is of radius 50 (similar to the wind speed) and the larger circle is of radius 200 (similar to the plane vector.  To get the coordinates of these two vectors,  use polar coordinates.

Let East be 0 degrees, so since the wind vector is on the circle of radius 50, we have:

w⃗ =⟨50cos(135),50sin(135)⟩=⟨−252√, 252√⟩.

P⃗ =⟨200cos(60),200sin(60)⟩=⟨100,\1003√⟩.

w⃗ +P⃗ =⟨100−252√ , 1003√+252√⟩

||w⃗ +P⃗ ||=(100−252√)2+(1003√+252√)2

√≈218.349218.

good looking response

b. lifting dumbbell weight at 20-degree angle from the body after a 45-degree angle